Question

Let \( A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \) and \[B = I + \text{adj}(A) + (\text{adj}(A))^2 + \dots + (\text{adj}(A))^{10}.\]Then, the sum of all the elements of the matrix \( B \) is:

• A. -110
• B. 22
• C. -88
• D. -124

Answer 1

The Correct Option is C

Given \( A = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \). The adjugate of \( A \), denoted as \(\text{adj}(A)\), is the transpose of its cofactor matrix.

First, \(\text{adj}(A)\) is calculated as:

\(\text{adj}(A) = \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}\)

Next, we compute \(\text{adj}(A)^2\):

\(\text{adj}(A)^2 = \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -4 \\ 0 & 1 \end{pmatrix}\)

Subsequently, \(\text{adj}(A)^{10}\) is determined by continuing this pattern:

\(\text{adj}(A)^{10} = \begin{pmatrix} 1 & -20 \\ 0 & 1 \end{pmatrix}\)

The matrix \( B \) is the sum of the identity matrix and the powers of \(\text{adj}(A)\) up to the 10th power:

\(B = I + \text{adj}(A) + \text{adj}(A)^2 + \cdots + \text{adj}(A)^{10}\)

\(B = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix} + \begin{pmatrix} 1 & -4 \\ 0 & 1 \end{pmatrix} + \cdots + \begin{pmatrix} 1 & -20 \\ 0 & 1 \end{pmatrix}\)

The resulting matrix \( B \) is:

\(B = \begin{pmatrix} 11 & -110 \\ 0 & 11 \end{pmatrix}\)

The sum of all elements in \( B \) is:

\(11 + (-110) + 11 = -88\)